# Heating and Cooling down time of an object

Let's say I have 2 identical metal bars, same properties and everything, both are at 100 degrees Celsius. Is the time heating up one metal bar to 120 the same as cooling one down to 80 degree Celcius?

• It depends on how you heat and how you cool.
– V.F.
Apr 12, 2018 at 4:42
• Assuming the methods of heating and cooling are the same and in ideal conditions (no heat transfer within the materials, no heat loss)
– DKHN
Apr 12, 2018 at 6:29
• In this case, loosely speaking, yes. For instance, if for cooling you use an identical bar at 60 degrees and for heating - an identical bar at 140 degrees.
– V.F.
Apr 12, 2018 at 13:26
• Are we acting within the linear range of the material properties?
– JMac
Apr 12, 2018 at 18:00

I think we can assume that it is, formula for heating/cooling of metals is $$Q=m*c*\Delta t$$ where $t$ is temperature, as $\Delta t$'s are equal, same with required heat and time if we use same power in both cases.

• The same amount of heat being required does not mean that the heating time is equal to the cooling time. Apr 12, 2018 at 11:38
• I think at the same power usage time will be also similar Apr 12, 2018 at 13:58
• In cooling, what is the power usage? Apr 12, 2018 at 14:39

Expanding on the comments...

One of the ways to figure that out without solving equations is to take advantage of free and easy to use electrical spice programs and model heat transfer with electrical circuits. This should work because, with a proper modeling, since underlying equations for charge and heat transfer are similar.

Let's say, we want to set up symmetrical cooling and heating conditions (as proposed in the comments), using identical bars as cooling and heating bodies.

The metal bar at 100 degrees could be modeled as a capacitor, say, 1uF charged to, say, 100V.

The cooling bar with the initial temperature of 60 degrees could be modeled by a 1uF cap charged to 60V. Similarly, the heating bar at 140 degrees could be modeled by a 1uF capacitor charged to 140V.

The thermal resistance of the thermal contact between the bars could be modeled as an electrical resistance between the capacitors, presumably, the same for both cooling and heating scenarios, say, 1Mohm.

The circuits and the simulation results are shown below. Here we can see that, due to symmetrical conditions and linearity, the cooling and heating will take the same time.

NOTE: Since we don't know the actual physical characteristics of the bars, we are picking random electrical equivalents and, correspondingly, the results of the simulation should not be interpreted as absolute.

If, for cooling and heating, we chose massive bodies which would maintain constant temperatures throughout the process, we could model them as perfect voltage sources of 80V and 120V, where 80V source would act as a heat sink and 120V source would act as a heat source.

Here, again, due to symmetrical conditions and linearity, the heat transfer will take the same time.